1. **State the problem:** We are given a triangle with angle $\theta$ such that $\cos \theta = \frac{1}{8}$. The triangle's sides around angle $\theta$ are labeled $x+3$, $4x$, and the side opposite $\theta$ is $y$. 2. **Recall the Law of Cosines:** For any triangle with sides $a$, $b$, and opposite side $c$ to angle $\theta$, the law states: $$ c^2 = a^2 + b^2 - 2ab\cos \theta $$ Here, $a = x+3$, $b = 4x$, and $c = y$. 3. **Apply the Law of Cosines:** Substitute the values into the formula: $$ y^2 = (x+3)^2 + (4x)^2 - 2(x+3)(4x)\frac{1}{8} $$ 4. **Simplify each term:** $$(x+3)^2 = x^2 + 6x + 9$$ $$(4x)^2 = 16x^2$$ $$- 2(x+3)(4x)\frac{1}{8} = - \frac{2 \times 4x (x+3)}{8} = - \frac{8x(x+3)}{8} = -x(x+3) = - (x^2 + 3x)$$ 5. **Put the simplified terms together:** $$ y^2 = x^2 + 6x + 9 + 16x^2 - (x^2 + 3x) $$ $$ y^2 = (x^2 + 16x^2 - x^2) + (6x - 3x) + 9 = 16x^2 + 3x + 9 $$ 6. **Identify the coefficients:** $$a = 16, \quad b = 3, \quad c = 9$$